RI Electrochemistry (I) Tutorial Answers

6 ANSWERS TO SELF-CHECK QUESTIONS 1 (a) The standard electrode potential, E, of a half–cell is the electromotive force, measured at 298 K, between the half–cell and the standard hydrogen electrode, in which the concentration of any reacting species in solution is 1 mol dm–3 and any gaseous species is at a pressure of 1 bar. (b) The standard cell potential (Ecell) is the potential difference between two half–cells under standard conditions. 2 (a) Checklist for the general labels to include when drawing the setup: in each half-cell connecting 2 half-cells  identity of electrode  external wire with voltmeter  identity and concentration of electrolyte  direction of electron flow  gas jar, identity and partial pressure of gas (if any)  salt bridge  temperature Note: Question may also ask for the labelling of anode (–) or cathode (+). (b) Set up the electrochemical cell as shown above. Read off the Ecell value on the voltmeter. The greater the Ecell value, the greater the difference in electrode potentials of the two half-cells. Monitor the direction of the flow of electrons to identify the electrode that is positively charged (the one to which electrons are flowing towards). The electrode that is positively charged contains the stronger oxidising agent which is preferentially reduced. (c) Cl2 (g) + 2I (aq)  I2 (aq) + 2Cl(aq) 3 (a) reduction half cell: Cl2 / Cl (b) Cl2 + Co  2Cl + Co2+ (c) Ecell = +1.36 – (–0.28) = +1.64 V

7 ANSWERS TO PRACTICE QUESTIONS 4 Cell (1) (a) (i) Cr 2O72–(aq) + 14H+(aq) + 6e ⇌ 2Cr3+(aq) + 7H2O(l) E = +1.33 V ... (1) Cu 2+(aq) + 2e ⇌ Cu(s) E = +0.34 V ... (2) Reduction: Cr2O72– (aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7H2O(l) Cathode: Pt electrode in Cr2O72–(aq) / Cr3+(aq) half–cell Oxidation: Cu(s)  Cu2+(aq) + 2e Anode: Copper electrode in Cu2+(aq) / Cu(s) half–cell Overall: Cr2O72–(aq) + 14H+(aq) + 3Cu(s)  2Cr3+(aq) + 7H2O(l) + 3Cu2+(aq) Note: Students are to note that both the half and full equations are written with  instead of ⇌ seen in the Data Booklet. (ii) Ecell = Ecathode – Eanode = +1.33 – (+0.34) = +0.99 V (iii) Cell (2) (a) (i) H 2O2(aq) + 2H+(aq) + 2e ⇌ 2H2O(l) E = +1.77 V O2(g) + 2H+(aq) + 2e ⇌ H2O2(aq) E = +0.68 V Reduction: H2O2(aq) + 2H+(aq) + 2e  2H2O(l) Cathode: Pt electrode in H2O2(aq) / H2O(l) half–cell Oxidation: H2O2(aq)  O2(g) + 2H+(aq) + 2e Anode: Pt electrode in O2(g) / H2O2(aq) half–cell Overall: 2H2O2(aq)  2H2O(l) + O2(g) Note: Students are to note that both the half and full equations are written with  instead of ⇌ seen in the Data Booklet. (iii) Ecell = Ecathode – Eanode = +1.77 – (+0.68) = +1.09 V V T = 298 K Pt Cu salt bridge [Cu2+(aq)] = 1 mol dm–3 [Cr2O72– (aq)] = [Cr3+(aq)] = [H+(aq)] = 1 mol dm–3 e–

8 (iv) (b) (i) Cu2+ precipitates out as CuCO3 and so [Cu2+(aq)] decreases. This shifts the position of equilibrium of (2) to the left so that E(Cu2+/Cu) becomes less positive. Hence, Ecell becomes more positive (i.e. Ecell > Ecell). (ii) Br2 + 2e– ⇌ 2Br– E = +1.07 V Br– reduces Cr2O72 to Cr3+. Note: Ecell = +1.33 – (+1.07) = +0.26 V > 0 (feasible) This decreases [Cr2O72(aq)] but increases [Cr3+(aq)] so that the position of equilibrium of (1) shifts to the left and so E(Cr2O72/Cr3+) becomes less positive. Hence Ecell becomes less positive (i.e. Ecell < Ecell). (c) To decrease Ecell, either (1) decrease E(H2O2/H2O) or (2) increase E(O2/H2O2). Change (1): E(H2O2/H2O) can be decreased by using H2O2 or H+ of a lower concentration (i.e. < 1 mol dm–3) the H2O2/H2O half–cell. Change (2): E(O2/H2O2) can be increased by using O2 of a higher pressure (i.e. > 1 bar), H+ of a higher concentration (i.e. > 1 mol dm–3) or H2O2 of a lower concentration (i.e. < 1 mol dm–3) in the O2 / H2O2 half–cell. 5 From Data Booklet, E(H+/H2) = 0.00 V; E(Ag+/Ag) = +0.80 V In the reaction 2H+(aq) + Cd(s)  Cd2+(aq) + H2(g), Cd is oxidised. Thus Ecell = Ecathode – Eanode = E(H+/H2) – E(Cd2+/Cd) = 0.00 – E(Cd2+/Cd) = +0.40 V  E(Cd2+/Cd) = –0.40 V In the reaction Pd2+(aq) + 2Ag(s)  2Ag+(aq) + Pd(s), Pd2+ is reduced. Thus Ecell = Ecathode – Eanode = E(Pd2+/Pd) – E(Ag+/Ag) = E(Pd2+/Pd) – 0.80 = +0.19 V  E(Pd2+/Pd) = +0.99 V Therefore, potential of Pd relative to Cd is +0.99 – (–0.40) = +1.39 V V T = 298 K Pt [H2O2(aq)] = [H+(aq)] = 1 mol dm–3 Pt O2 (1 bar) [H2O2(aq)] = [H+(aq)] = 1 mol dm–3 salt bridge e– Note to tutors: Highlight during the lesson that the gas (@1 bar) may be the reactant or product – i.e. the gas must be present so long as it appears in the half-equation. (common misconception for voltaic cell is that the gas @ 1 bar is only required if it is the reactant)

9 6 (a) (i) O2 + 4H+ + 4e  2H2O Note: This half-equation can be found in the Data Booklet. (ii) Reduction: O2 + 4H+ + 4e  2H2O ( 3) Oxidation: CH3OH + H2O  CO2 + 6H+ + 6e ( 2) Overall: 2CH3OH + 3O2  4H2O + 2CO2 (iii) From Data Booklet, E(O2/H2O) = +1.23 V Ecell = Ecathode – Eanode Thus, +1.18 = +1.23  E(CO2/CH3OH) E(CO2/CH3OH) = +1.23  (+1.18) = +0.05 V (iv) Any one of the following:  Methanol has a more efficient storage (by volume or by weight since it is a liquid) compared to compressed hydrogen gas.  Hydrogen gas is more volatile and explosive, thus requires high pressure storage system but not methanol (less explosive). (b) (i) CH3CH2OH + 3O2  2CO2 + 3H2O (ii) G = H  TS G = (1367  103)  (298)(140) = 1.33 106 J mol1 = 1330 kJ mol1 (iii) Ecell = = +1.15 V 7 (a) (i) A saturated KCl solution is used to keep the concentration Cl− constant in the reference electrode. Examiner Comments  This question was poorly done. The hint was given in the question regarding the nature of a reference electrode (fixed composition and constant potential). (ii) The E value is more positive than +0.241 V. At standard conditions, the KCl concentration is 1.00 mol dm3, which is lower than the saturated KCl solution. Therefore, the equilibrium position of the half-equation lies more to the right, resulting in a more positive E value. Note: Concentration of a saturated solution of KCl is 4.55 mol dm–3 at 20 ºC. Examiner Comments  Students who lost marks here were careless in their reading of the question. The E value (+0.241 V) given is not at standard conditions.  When answering questions, students should be clear in referring to E or E. E is used to represent the potential at any conditions that are not standard. (b) (i) Ag+(aq) + e ⇌ Ag(s) E = +0.80 V Since the Ag+/Ag electrode has the more positive E, it is the cathode. The S.C.E. is the anode.   6 -1 -1 ( 1.33 × 10 J mol ) 12 (96500 C mol )

10 (ii) Ecell = +0.80 – 0.241 = +0.559 V Examiner Comments  This question was generally well done.  Students are reminded to express their answers in 3 s.f. and with units. (iii) At very low concentrations of Ag+, small increases in [Ag+] result in an exponential/ large increase in Ecell as shown by the steep gradient of the graph. Examiner Comments  Some students described the shape of the graph in terms of rate, e.g. increased rapidly, increased at a greater rate, etc. This is not accepted as there is no time scale on the graph.  A number of students discussed the gentle gradient at relatively higher [Ag+]. This is not accepted as it doesn’t answer the question. 8 (a) Ecell = +0.80 – (+0.34) = +0.46 V (spontaneous) (b) Ecell = +1.07 – (+0.77) = +0.30 V (spontaneous) (c) Ecell = –1.66 – (+1.36) = –3.02 V (not spontaneous)